> For example, if cell G(1,1) is active for a location on an object, a movement 
> one step to the right will cause G(2,1) to become active. Since the cells 
> tile space, an additional movement one step to the right will cause G(0,1) to 
> become active.

![Figure%20A|388x465](upload://ouT6j3mB7WA7geIIwIcrJNOSMDa.png) 

The supplementary material covers single step movements. In the Grid cell 
module in the **Figure A** when the cell G(1,1) is active a single step 
movement can be 1 step to the "left", "right", "up", "down", "left-up", 
"left-down", "right-up", "right-down" (diagonally) or "none" (same place but 1 
step through time). All these count for 9 relative displacements of 1 step. 
That's why there are 9 Displacement cells representing them in the Displacement 
module.

![Figure%20B|307x451](upload://3vdkUpG6wztSYp8ECRhRNo9MXsF.png) 

**Figure B** shows that independent of which cell performs the 1 step to the 
right movement D(1,0) will represent the displacement. The formula in the 
supplementary material for calculating which displacement cell will become 
active is:

G(x1,y1) --> G(x2,y2): D((x2-x1) mod 3, (y2-y1) mod 3)

Testing with 1 step to the right:

G(0,1) --> G(1,1): D(1 mod 3, 0 mod 3) = D(1,0)
G(1,1) --> G(2,1): D(1 mod 3, 0 mod 3) = D(1,0)
G(2,1) --> G(0,1): D(-2 mod 3, 0 mod 3) = D(1,0)

Testing with 2 steps to the right:

G(0,1) --> G(2,1): D(2 mod 3, 0 mod 3) = D(2,0)

But D(2,0) is representing 1 step to the left.

>From G(0,1) 1 step to the left or 2 steps to the right will both land on 
>G(2,1) but the displacement is different.

Questions:
1. Can there be displacement cells representing 2,3,n steps?
2. Is this problem solved more elegantly by combining grid+displacement cells 
made to work on a different scale?
4. Is it done by extending the capabilities of displacement cells by changing 
the anatomy to different types of cells like conjunctive cells?





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